Let $\zeta=\cos(\frac{2\pi}{16})+i\sin(\frac{2\pi}{16})$ be a 16th root of unity, so that it is a primitive root of unity. I need to explicitly express this number in terms of radicals: $a+ib$, where $a$ and $b$ are real numbers.
My approach:
We know that $\zeta^1+\zeta^2+...+\zeta^{16} = 0$. We can now compute $\zeta$ in even powers to see that they all cancel out and so we have:
$$\zeta^2+...+\zeta^{16}=0$$ Dividing by $\zeta^2$ we get:
$$ 1+\frac{1}{\zeta^2}+\zeta^2+\zeta^4+\zeta^6+\zeta^8+...+\zeta^{14}=0 $$ which we can rewrite as $$ 1+ \frac{1}{\zeta^2}+(\zeta^2+\zeta^4+\zeta^6)+\zeta^6(\zeta^2+\zeta^4+\zeta^6)=1+ \frac{1}{\zeta^2}+(1+\zeta^6)(\zeta^2+\zeta^4+\zeta^6)$$
Well, I don't think this is useful anyway. Would appreciate some hints. I'm not sure how to approach this problem in a right way.
Given that $\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$, and that $\cos (\frac{a}{2})=\pm \sqrt{\frac{1+\cos a}{2}}$ and $\sin (\frac{a}{2})=\pm \sqrt{\frac{1-\cos a}{2}}$, you find that $\cos \frac{\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$ and $\sin \frac{\pi}{8}=\frac{\sqrt{2-\sqrt{2}}}{2}$. That is,
$$\cos(\frac{2\pi}{16})+i\sin(\frac{2\pi}{16})=\frac{\sqrt{2+\sqrt{2}}}{2}+i\cdot\frac{\sqrt{2-\sqrt{2}}}{2}$$